4 pipes can fill a tank in 15, 20, 30 & 60 hours respectively. The first was opened at 6 a.m., second at 7 a.m., third at 8 a.m. and fourth at 9 a.m. When will the tank be full?

To solve this problem, we first need to find the rate at which each pipe fills the tank.

1. The first pipe can fill the tank in 15 hours, so it fills 1/15 of the tank per hour.
2. The second pipe can fill the tank in 20 hours, so it fills 1/20 of the tank per hour.
3. The third pipe can fill the tank in 30 hours, so it fills 1/30 of the tank per hour.
4. The fourth pipe can fill the tank in 60 hours, so it fills 1/60 of the tank per hour.

Next, we need to calculate how much of the tank each pipe fills by the time the fourth pipe is opened at 9 a.m.

1. The first pipe was opened at 6 a.m., so it has been open for 3 hours. In that time, it has filled 3 * (1/15) = 1/5 of the tank.
2. The second pipe was opened at 7 a.m., so it has been open for 2 hours. In that time, it has filled 2 * (1/20) = 1/10 of the tank.
3. The third pipe was opened at 8 a.m., so it has been open for 1 hour. In that time, it has filled 1 * (1/30) = 1/30 of the tank.
4. The fourth pipe has just been opened, so it has not yet filled any of the tank.

Adding these fractions together, we find that by 9 a.m., the pipes have filled 1/5 + 1/10 + 1/30 = 6/30 + 3/30 + 1/30 = 10/30 = 1/3 of the tank.

After 9 a.m., all four pipes are open, so they fill the tank at a combined rate of 1/15 + 1/20 + 1/30 + 1/60 = 4/60 + 3/60 + 2/60 + 1/60 = 10/60 = 1/6 of the tank per hour.

Since 1/3 of the tank is already full, there is 1 - 1/3 = 2/3 of the tank left to fill. At a rate of 1/6 of the tank per hour, this will take 2/3 ÷ 1/6 = 4 hours.

Therefore, the tank will be full at 9 a.m. + 4 hours = 1 p.m.